Question: Find $\sin\left(\dfrac{5\pi}{12}\right)$ exactly using an angle addition or subtraction formula. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{-\sqrt{2}-\sqrt{6}}{4}$ (Choice B) B $\dfrac{\sqrt{2}+\sqrt{6}}{4}$ (Choice C) C $\dfrac{\sqrt{3}+\sqrt{2}}{8}$ (Choice D) D $\dfrac{\sqrt{3}-1}{8}$
The strategy First, we should rewrite the given angle $\dfrac{5\pi}{12}$ as the sum or difference of two special angles. Then, we can use the sine addition or subtraction identities in order to evaluate $\sin\left(\dfrac{5\pi}{12}\right)$. [How do we find the trigonometric value of a sum or difference?] Rewriting $\dfrac{5\pi}{12}$ We can rewrite $\dfrac{5\pi}{12}$ as follows. $\begin{aligned}\dfrac{5\pi}{12}&=\dfrac{2\pi}{12}+\dfrac{3\pi}{12}\\\\\\ &=\dfrac{\pi}{6}+\dfrac{\pi}{4}\end{aligned}$ In other words, $\dfrac{5\pi}{12}$ is the sum of the special angles $\dfrac{\pi}{6}$ and $\dfrac{\pi}{4}$. Evaluating $\sin\left(\dfrac{5\pi}{12}\right)$ Using the sine addition identity, we get the following. $\begin{aligned} \sin\left(\dfrac{5\pi}{12}\right)&= \sin\left(\dfrac{\pi}{6}+\dfrac{\pi}{4}\right) \\\\\\ &= \sin \left(\dfrac{\pi}{6}\right) \cos \left(\dfrac{\pi}{4}\right) + \cos \left(\dfrac{\pi}{6}\right) \sin \left(\dfrac{\pi}{4}\right) \\\\\\ &=\left(\dfrac{1}{2}\right) \left(\dfrac{\sqrt{2}}{2}\right) + \left(\dfrac{\sqrt{3}}{2}\right) \left(\dfrac{\sqrt{2}}{2}\right) \\\\\\ &=\left(\dfrac{\sqrt{2}}{4}\right) + \left(\dfrac{\sqrt{6}}{4}\right)\\\\\\ &=\dfrac{\sqrt{2}+\sqrt{6}}{4} \end{aligned}$ Summary $\sin\left(\dfrac{5\pi}{12}\right) = \dfrac{\sqrt{2}+\sqrt{6}}{4}$